Basic Engineering Mathematics, Fifth Edition

70 Basic Engineering Mathematics

(i) What multipliesato makeab?Answer:b

(ii) What multipliesato make− 5 ac?Answer:− 5 c

Hence,b− 5 cappears in the bracket. Thus,

ab− 5 ac=a(b− 5 c)

Problem 10. Factorize 2x^2 + 14 xy^3

For the numbers 2 and 14, the highest common factor

(HCF) is 2 (i.e. 2 is the largest number that divides into

both 2 and 14).

For thexterms,x^2 andx,theHCFisx.

Thus, the HCF of 2x^2 and 14xy^3 is 2x.

2 xis therefore taken outside of the bracket. What goes

inside the bracket?

(i) What multiplies 2xto make 2x^2 ?Answer:x

(ii) What multiplies 2xto make 14xy^3 ?Answer:7y^3

Hencex+ 7 y^3 appears inside the bracket. Thus,

2 x^2 + 14 xy^3 = 2 x(x+ 7 y^3 )

Problem 11. Factorize 3x^3 y− 12 xy^2 + 15 xy

Forthenumbers3,12and15,thehighestcommonfactor

is 3 (i.e. 3 is the largest number that divides into 3, 12

and 15).

For thexterms,x^3 ,xandx,theHCFisx.

For theyterms,y,y^2 andy,theHCFisy.

Thus, the HCF of 3x^3 yand 12xy^2 and 15xyis 3xy.

3 xyis therefore taken outside of the bracket. What goes

inside the bracket?

(i) What multiplies 3xyto make 3x^3 y?Answer:x^2

(ii) What multiplies 3xyto make− 12 xy^2 ?Answer:

− 4 y

(iii) What multiplies 3xyto make 15xy?Answer:5

Hence,x^2 − 4 y+5 appears inside the bracket. Thus,

3 x^3 y− 12 xy^2 + 15 xy= 3 xy(x^2 − 4 y+ 5 )

Problem 12. Factorize 25a^2 b^5 − 5 a^3 b^2

For the numbers 25 and 5, the highest common factor

is 5 (i.e. 5 is the largest number that divides into 25

and 5).

For theaterms,a^2 anda^3 ,theHCFisa^2.

For thebterms,b^5 andb^2 ,theHCFisb^2.

Thus, the HCF of 25a^2 b^5 and 5a^3 b^2 is 5a^2 b^2.

5 a^2 b^2 is therefore taken outside of the bracket. What

goes inside the bracket?

(i) What multiplies 5a^2 b^2 to make 25a^2 b^5 ?Answer:

5 b^3

(ii) What multiplies5a^2 b^2 to make− 5 a^3 b^2 ?Answer:

−a

Hence, 5b^3 −aappears in the bracket. Thus,

25 a^2 b^5 − 5 a^3 b^2 = 5 a^2 b^2 ( 5 b^3 −a)

Problem 13. Factorizeax−ay+bx−by

The first two terms have a common factor ofaand the

last two terms a common factor ofb. Thus,

ax−ay+bx−by=a(x−y)+b(x−y)

The two newly formed terms have a common factor of

(x−y). Thus,

a(x−y)+b(x−y)=(x−y)(a+b)

Problem 14. Factorize 2ax− 3 ay+ 2 bx− 3 by

ais a common factor of the first two terms andba

common factor of the last two terms. Thus,

2 ax− 3 ay+ 2 bx− 3 by=a( 2 x− 3 y)+b( 2 x− 3 y)

( 2 x− 3 y)is now a common factor. Thus,

a( 2 x− 3 y)+b( 2 x− 3 y)=( 2 x− 3 y)(a+b)

Alternatively, 2x is a common factor of the original

first and third terms and− 3 yis a common factor of

the second and fourth terms. Thus,

2 ax− 3 ay+ 2 bx− 3 by= 2 x(a+b)− 3 y(a+b)

(a+b)is now a common factor. Thus,

2 x(a+b)− 3 y(a+b)=(a+b)( 2 x− 3 y)

as before

Problem 15. Factorizex^3 + 3 x^2 −x− 3

x^2 is a common factor of the first two terms. Thus,

x^3 + 3 x^2 −x− 3 =x^2 (x+ 3 )−x− 3

−1 is a common factor of the last two terms. Thus,

x^2 (x+ 3 )−x− 3 =x^2 (x+ 3 )− 1 (x+ 3 )